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Ceiling and Floor Formulation (Posted on 2013-03-04) Difficulty: 3 of 5
Formulate an algorithm for fast evaluation of:
Σj=1,...,n2 (floor (√j) + ceil (√j)), where n is a positive integer.

** ceil(x) is the least integer ≥ x and, floor(x) is the greatest integer ≤ x

No Solution Yet Submitted by K Sengupta    
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Solution Solution? | Comment 1 of 6
Would finding the sum as a function of n count as a fast algorithm?

S(n) = (4n-n)/3

The floors from each n-1 to n have a sum n(2n+1) wolfram alpha gives the full sum from 1 to n-1 as (4n-3n-n)/6
The ceilings from each n-1 to n have a sum n(2n-1) wolfram alpha gives the full sum from 1 to n as (4n+3n-n)/6

Then just sum the parts.

  Posted by Jer on 2013-03-04 14:46:40
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